Find the jordan basis for $A$

Question

$$A =\left(\begin{array}{cc}5 & -4 \\9 & -7\end{array}\right)$$

I found that the eigenvalues are $-1$ (algebriac multiplicity 2)

Therefore, the jordan form looks like this:

$$J =\left(\begin{array}{cc}-1 & 1 \\0 & -1\end{array}\right)$$

Also solving $(A-\lambda I$)$=\vec{0}$, when $\lambda = -1$ gives us the nullspace $(3,2)^T$.

However I need one more basis vector, and unsure how to find it?

Answer 1

from Jordan theorem we know that a matrix M exists such that $$M^{-1}AM=J$$

let $$M=[v,w]$$

then M has to satisfy the following system: $$AM=MJ$$ that is in your case $$Av=-v$$ $$Aw=v-w$$ once you find $v$ from the first equation you can find also $w$ from the second.

Answer 2

You can take any vector which is linearly independent of the eigenvector, except that you need to adapt its length in order to get exactly $1$ (and not some other constant) in the upper right corner of $J$. For example, take something simple like $(k,0)^T$, see what matrix you get in the new basis, and then choose $k$ so that you get exactly $J$.